# Convolution of measures - entropy growth

Imagine you have two shift-invariant measures $\mu, \nu$ in the Bernoulli space $\{0,1\}^{\mathbb{N}}$ with positive entropy and both are not the Bernoulli measure $(\frac{1}{2},\frac{1}{2})$. I know that:

$$H( \mu \ast \nu) \geq \max(H( \mu ),H( \nu))$$

Question. How can I prove this result with $>$ instead of $\geq$?

• Do we have the inequality $H(\mu *\nu) \geq \max (H(\mu),H(\nu))$ for any positive measures in a compact group $G$ ? – BigBill Oct 18 '20 at 6:21
Suppose that $p\in[0,1]$ and call by $\mu^{(p)}$ the $(p,1-p)$-Bernoulli measure. Let $0\leq q,p\leq 1.$ Then $\mu^{(p)}*\mu^{(q)}=\mu^{(pq+(1-p)(1-q))}.$ In particular, if $q=0.5$ (or $p=0.5$) then $\mu^{(p)}*\mu^{(q)}=\mu^{(0.5)},$ and if $q=0,$ then $\mu^{(p)}*\mu^{(q)}=\mu^{(p)}$ (this is always true, but nice to check in this example). Using this observation, suppose that $0<q<p<1$ and $q,p\neq 0.5,$ then (an easy calculation gives that) $$H(\mu^{(p)}*\mu^{(q)})>\max\{H(\mu^{(p)}),H(\mu^{(q)})\}.$$
It is fun to consider $\epsilon>0$ small, and consider the function $$[0,1]\ni p\mapsto F(p):=H(\mu^{(p)}*\mu^{(\epsilon)}).$$ It has a unique maximum at $p=0.5$ with $F(0.5)=\log 2.$ It is monotone increasing on $[0,0.5],$ monotone decreasing on $[0.5,1]$ and $F(0)=F(1)=H(\mu^{(\epsilon)}).$